Quinic acid, A, has the gross structure shown below:

(i) Draw all the possible diastereoisomers of A and indicate which are chiral.
(ii) Quinic acid is optically active and the marked hydroxyl groups are cis to one another; which of your structures is consistent with this data?

Using the $\ce{^{1}H}$-NMR data below, draw a likely conformation for quinic acid, making sure that you fully assign all these data.

$\ce{^{1}H}$-NMR for quinic acid in $\ce{D2O}$: δ 4.09 (1H, q, 3 Hz), 3.99 (1H, ddd, 11 Hz, 9 Hz, 5 Hz), 3.39 (1H, dd, 3 Hz, 9 Hz), 2.12 (1H, dd, 13 Hz, 5 Hz), 2.05 (2H, d, 3 Hz), 1.86 (1H, dd, 13 Hz, 11 Hz).

My working so far:

Firstly, have I missed any diastereoisomers? And what does it mean by 'conformation' of quinic acid? Does it mean a chair conformer?

But I'm not sure about the shift at 4.09 ppm since I made the assumption that the quartet is actually a doublet of doublets. Also surely $\ce{H_{f}}$ would vicinal couple with $\ce{OH}$ on the same carbon?

  • $\begingroup$ Also, you write that you only need the deduction for the last proton … so how did you deduce the placement of the other six? (This is a homework question which means you should show as much work as possible ;).) $\endgroup$
    – Jan
    Commented Dec 22, 2015 at 0:58
  • $\begingroup$ Here’s a hint for the diastereomers: Grab one group and bind it in place. I suggest the carbon attached to $\ce{COOH}$. Then, wobble the remaining ones back and forth. Since one stereocentre will not budge, you can be sure to not generate any enantiomers/identical structures. $\endgroup$
    – Jan
    Commented Dec 22, 2015 at 2:01
  • $\begingroup$ The last comment was supposed to mean: the structure which solves it is missing. If the quartet really had been a dd, they would have reported it as such; and if the dd had fallen together due to identical coupling constants, it would have fallen together into a triplet, not a quartet. I’ll add that to my answer. $\endgroup$
    – Jan
    Commented Dec 22, 2015 at 2:12
  • 1
    $\begingroup$ Confirming that if I can see it correctly in that minitiature image of yours, you got the diastereomers correct ^^ $\endgroup$
    – Jan
    Commented Dec 22, 2015 at 2:31

1 Answer 1


I see one major issue in your working immediately: The NMR has been recorded in $\ce{D2O}$, hence there will be no more $\ce{OH}$ groups present in your molecule, only $\ce{OD}$ groups. Since deuterium is not hydrogen, you cannot observe any coupling to the $\ce{OD}$ groups as if they contained hydrogen.

Also, you said:

I made the assumption that the quartet is actually a doublet of doublets.

This is an incorrect assumption. They have multiple dd’s that they reported as dd which gives us less reason to suspect an incorrect report. Note that typical dd signals have two different coupling constants, e.g. the proton at 2.12. Now you could imagine that the two doublets fall together. But in that case, they would fall together to become a triplet, not a quartet. You can try it with splitting trees: Either a dd stays dd, or it collapses into a t. And finally, the quartet cannot be a set of two doublet signals from distinct hydrogens since it only has an integral of 1.

That said, the attribution of different hydrogens to their position in the ring is remarkably simple. Let’s label the quaternary carbon $\mathrm{C1}$ and then go around the ring in any direction. Our first stop is $\mathrm{C2}$.

Since this is a $\ce{CH2}$ group, its chemical shift must be lower than $3~\mathrm{ppm}$. We have two choices. Either we take the two hydrogens that exhibit a $^2J$ coupling or the other two. Since we can easily identify the $^2J$ coupling, let’s take those.

With the hydrogens of $\mathrm{C2}$ in hand, we can check, which hydrogen must be on $\mathrm{C3}$.

Again it’s pretty simple. The two coupling constants that are not $^2J$ constants must all be present in the hydrogen’s signal. There is only one in question.

Once we have gotten that sorted, we can go back and determine the stereochemistry that must be present here.

We have one very large $^3J$ coupling and one mediocre one. The very large coupling is, of course, due to an axial-axial relationship between the two hydrogens.

We now have three hydrogens firmly in place: Those on $\mathrm{C2}$ and the one on $\mathrm{C3}$. Let’s move on to $\mathrm{C4}$.

There is one coupling constant on the hydrogen at $\mathrm{C3}$ which is not yet assigned and which thus must belong to the coupling to this hydrogen. It’s magnitude again tells us pretty clearly which dihedral angle we expect and thus what the relationship between the two must be.

With $\mathrm{C4}$’s stereochemistry sorted out, $\mathrm{C5}$ is next.

The coupling constant is rather different this time, indicating the opposite relative stereochemistry. Note that the hydrogen at $\mathrm{C5}$ is reported as a quartet of the same coupling. We’ll get back to that in a second. For now, we know what must be substantially different here.

And now we can move on to $\mathrm{C6}$, the one carbon we are still missing.

As I just mentioned, we have an apparant quartet for the hydrogen on $\mathrm{C5}$. But that’s okay. It just means that there are three rather similar coupling constants to neighbouring hydrogens. One is taken from the coupling to $\mathrm{C4}$’s hydrogen, meaning the two of $\mathrm{C6}$ must be the other two. Thankfully, there is one signal with an integral of two left, which also has the same coupling constant. And also thankfully, if you did everything correctly, there should be no questions as to why this stereochemistry along the $\ce{C{5}-C{6}}$ bond be correct.

Finally, we have a hint in the question as to the overall structure:

Namely that the two hydrogens on $\mathrm{C1}$ and $\mathrm{C4}$ are syn. Thus, the carboxylic acid must be in equatorial position and the hydroxy group axial.

A note on finding the diastereomers: Enantiomers are only enantiomers if all stereocentres are reversed. Thus, if you take one centre and keep it the same all the time and only exchange the direction of the others, you are bound to find diastereomers only. I suggest immobilising the $\mathrm{C1}$ carbon since it is rather prominent up there all alone; it gives the entire thing less hassle. Also, any achiral isomers will always have a plane of symmetry through it which helps, too.


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